Key Concepts and Formulas

• A complex number $z$ can be represented as $z = x + iy$, where $x = \operatorname{Re}(z)$ and $y = \operatorname{Im}(z)$.
• For two complex numbers $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$:
  • $z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$
  • $z_1 z_2 = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1)$
• $\operatorname{Re}(z_1 + z_2) = \operatorname{Re}(z_1) + \operatorname{Re}(z_2)$ and $\operatorname{Re}(z_1 z_2) = x_1x_2 - y_1y_2$.

Step-by-Step Solution

1. Represent $z_1$ and $z_2$ in algebraic form

Let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$, where $x_1, y_1, x_2, y_2$ are real numbers. This allows us to work with the real and imaginary parts separately.

2. Utilize the condition $\operatorname{Re}(z_1 + z_2) = 0$

We have $z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$. The condition $\operatorname{Re}(z_1 + z_2) = 0$ implies: $x_1 + x_2 = 0$ $x_2 = -x_1 \quad \text{(Equation 1)}$ This equation shows a relationship between the real parts of the two complex numbers.

3. Utilize the condition $\operatorname{Re}(z_1 z_2) = 0$

We have $z_1 z_2 = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1)$. The condition $\operatorname{Re}(z_1 z_2) = 0$ implies: $x_1x_2 - y_1y_2 = 0 \quad \text{(Equation 2)}$ This equation relates the real and imaginary parts of both complex numbers.

4. Combine the derived relationships

Substitute $x_2 = -x_1$ from Equation 1 into Equation 2: $x_1(-x_1) - y_1y_2 = 0$ $-x_1^2 - y_1y_2 = 0$ $y_1y_2 = -x_1^2 \quad \text{(Equation 3)}$ This equation simplifies the relationship and eliminates $x_2$.

5. Analyze the implications of $y_1y_2 = -x_1^2$

Since $z_1$ and $z_2$ are non-zero complex numbers, $x_1$ cannot be zero. If $x_1=0$, then $x_2 = -x_1 = 0$. If both $x_1$ and $x_2$ are zero, then $y_1$ and $y_2$ cannot both be zero, but equation 3 would then yield $y_1y_2 = 0$, so one of them must be zero. That makes either $z_1$ or $z_2$ zero, which contradicts the problem statement. Therefore $x_1 \neq 0$. Thus, $x_1^2 > 0$, which means $-x_1^2 < 0$. From Equation 3, we have $y_1y_2 = -x_1^2 < 0$.

6. Interpret $y_1y_2 < 0$ in terms of imaginary parts

The product $y_1y_2$ is negative if and only if $y_1$ and $y_2$ have opposite signs. This means:

• Either $y_1 > 0$ and $y_2 < 0$, which means $\operatorname{Im}(z_1) > 0$ and $\operatorname{Im}(z_2) < 0$. This corresponds to option C.
• Or $y_1 < 0$ and $y_2 > 0$, which means $\operatorname{Im}(z_1) < 0$ and $\operatorname{Im}(z_2) > 0$. This corresponds to option B.

7. Conclude the correct answer

Both possibilities B and C are valid. Therefore, the correct answer is the option that includes both B and C.

Common Mistakes & Tips

• Non-zero condition: Remember that $z_1$ and $z_2$ are non-zero. This implies that $x_1$ cannot be zero, which is crucial for the inequality $y_1y_2 < 0$.
• Sign errors: Be careful with signs, especially when substituting and simplifying equations. A small sign error can lead to an incorrect conclusion.
• Real vs. Imaginary: Clearly distinguish between real and imaginary parts when performing operations.

Summary

Given two non-zero complex numbers $z_1$ and $z_2$ such that $\operatorname{Re}(z_1 + z_2) = 0$ and $\operatorname{Re}(z_1 z_2) = 0$, we found that their real parts are additive inverses and their imaginary parts have opposite signs. This means either $\operatorname{Im}(z_1) > 0$ and $\operatorname{Im}(z_2) < 0$, or $\operatorname{Im}(z_1) < 0$ and $\operatorname{Im}(z_2) > 0$.

The final answer is \boxed{A}. which corresponds to option (A).